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General method of solid angle calculation using attitude kinematics 1

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Russell P. Patera1 3

351 Meredith Way, Titusville, FL 32780, United States 4

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Abstract 6

A general method to compute solid angle is developed that is based on Ishlinskii’s theorem, which specifies the 7 relationship between the attitude transformation of an axis that completely slews about a conical region and the 8 solid angle of the enclosed region. After an axis slews about a conical region and returns to its initial orientation, it 9 will have rotated by an angle precisely equal to the enclosed solid angle. The rotation is the magnitude of the 10 Euler rotation vector of the attitude transformation produced by the slewing motion. Therefore, the solid angle 11 can be computed by first computing the attitude transformation of an axis that slews about the solid angle region 12 and then computing the solid angle from the attitude transformation. This general method to compute the solid 13 angle involves approximating the solid angle region’s perimeter as seen from the source, with a discrete set of 14 points on the unit sphere, which join a set of great circle arcs that approximate the perimeter of the region. Pivot 15 Parameter methodology uses the defining set of points to compute the attitude transformation of the axis due to 16 its slewing motion about the enclosed solid angle region. The solid angle is the magnitude of the resulting Euler 17 rotation vector representing the transformation. The method was demonstrated by comparing results to 18 published results involving the solid angles of a circular disk radiation detector with respect to point, line and disk 19 shaped radiation sources. The proposed general method can be applied to any detector-source geometric 20 configuration. 21

Keywords: solid angle, geometrical efficiency, Ishlinskii’s theorem, Attitude kinematics, radiation source-detector 22 geometry 23

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Nomenclature 25

A semi-major axis of ellipse 26 a(k) x-component 27 B semi-minor axis of ellipse 28 b(k) y-component 29 dL increment of line source 30 d shift distance of parallel disk axis 31 G(k) solid angle of disk detector at kth point 32 GE geometrical efficiency 33 G1 solid angle of 1st Gaussian integration point 34 G2 solid angle of 2nd Gaussian integration point 35 h height of source above detector plane 36 I integration of solid angle over region 37 IE integration of solid angle over enlarged region 38 IS integration of solid angle over source disk 39 L1 beginning of line source 40 L2 end of line source 41 R radius of detector 42

1 Russell.p.patera@gmail.com

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R0 radius of expanded detector 1 R(r,α) rotational transformation about r by α 2 r(m) midpoint of line segment 3 S1 1st point of Gaussian integration 4 S2 2nd pint of Gaussian integration 5 S(k) kth point of line source 6 S source location 7 sin1 sine of 1st Gaussian angle 8 sin2 sine of 2nd Gaussian angle 9 U(a,b) slewing transformation from a to b 10 V Euler rotation vector 11 x(m) x coordinate of midpoint of line segment 12 y(m) y coordinate of midpoint of line segment 13 Ω solid angle 14 ΩS solid angle of detector seen from source 15 ΩR solid angle of source seen from detector 16 Φ polar angle 17 Φ1 1st Gaussian integration angle 18 Φ2 2nd Gaussian integration angle 19 ρ radius of disk source 20 Ψ(a) R(a,180) 21 ϴ polar angle 22 ϴ1 1st Gaussian integration angle 23 ϴ2 2nd Gaussian integration angle 24

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1. Introduction 26

The need to evaluate the solid angle appears in very diverse areas ranging from radiant heat transfer, particle 27 scattering cross sections, electric and magnetic field strength around charge distributions, evaluating the size of 28 ligands in metal complexes, and calibrating the strength of radioisotopes by computing the geometrical efficiency 29 of various source and detector geometries. The methods used to compute solid angle start with the fundamental 30 integral equation and employ various mathematical methods to compute the integral. Paxton (1959), found a 31 general expression for the solid angle subtended by a circular disk in terms of complete elliptic integrals of the first 32 and third kinds. Gardner and Carnesale (1969), found the solid angle subtended by a circular disk by replacing the 33 disc with a square of equal area and using a tabular correction factor. Gardner and Verghese (1971), found the 34 solid angle subtended by a circular disc using the analytical expression for the solid angle for an n-sided regular 35 polygon of area equal to the circular disk. Timus et. al. (2007), found a series formulation involving complete 36 elliptic integrals for the first and second kinds for the solid angle subtended at a point by a circular disk. Conway 37 (2010a), found an analytical solution for the solid angle subtended at any point by an ellipse using a point source 38 radiation vector potential. Thabet, et. al. (2020), found the solid angle subtended by an ellipse with respect to a 39 point on its axis. Conway (2010b), found the geometric efficiency for a circular detector and a linear source of 40 arbitrary orientation and position. Abbas (2018), determined the solid angle and geometrical efficiency for a 41 circular detector and an arbitrarily positioned line source. Ruby and Rechen (1968), computed the solid angle 42 subtended at a disk source by a coaxial parallel-disk detector using Bessel functions of order unity. Conway (2006), 43 generalized Ruby and Rechen’s method (1968), to include cases where he detector disk and source disk are parallel 44 but not necessarily coaxial. Thabet, et. al. (2020), computed the solid angle of a point source and a disk detector 45 by transforming to axial geometry with an equivalent elliptical detector cross section. They used the results to 46 evaluate the average geometrical efficiency of a coaxial disk source subtended by a circular detector using elliptical 47 integrals. All of the methods mentioned above made use of simple geometric relationships between source and 48 detector, as well as, various related special functions and integrals. 49 50

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In this work, a general method is developed to compute the solid angle for arbitrary source and detector geometry 1 without use of special functions or integrals. Instead of computing the area on the unit sphere to obtain the solid 2 angle, the method introduced in this work evaluates the attitude transformation associated with an axis centered 3 at the origin that sweeps around the perimeter of the enclosed area on the unit sphere. This method is based on 4 Ishlinskii’s theorem (1952), which states, if an axis fixed in the body, describes a closed conical surface in space, the 5 angle of rotation of the body around this axis is equal to the integral of the projection of the angular velocity of the 6 body onto this axis plus the solid angle of the cone enclosed. If the angular rate is zero, the rotation about the axis 7 after it completes one circuit about the perimeter is equal to the solid angle of the enclosed cone. The attitude 8 transformation produced by the slewing motion of the axis about the enclosed cone can be represented by a Euler 9 rotation vector aligned with the initial orientations of the axis, which is also the final orientation of the axis. The 10 magnitude of the Euler rotation vector is the rotation about the axis and desired solid angle of the enclosed region. 11 Therefore, the solid angle for an arbitrary region on the surface of a unit sphere can be found by first computing 12 the attitude transformation of the axis centered at the origin that slews about the region. The rotation angle can 13 then be obtained from the attitude transformation. The transformation can be represented by a direction cosine 14 matrix or a Euler rotation vector. The associated rotation angle, which is the desired solid angle, can be extracted 15 directly from the direction cosine matrix or can be obtained as the magnitude of the Euler Rotation vector 16 representing the transformation. The attitude transformation of the slewing motion of the axis can be found using 17 Pivot Parameters, Patera (2017). Pivot Parameters are particularly effective in computing slewing transformations 18 that can be viewed as great circle arcs on the unit sphere. The use of Pivot Parameters in computing solid angles 19 of closed regions bounded by great circle arcs on the unit sphere has been presented in a related work on attitude 20 kinematics, Patera (2020). 21 22 Section 2 presents the method to compute the attitude transformation and associated solid angle for an arbitrary 23 region on the unit sphere bounded by great circle arcs using the Pivot Parameter methodology. The attitude 24 transformation is in the form of a direction cosine matrix and a simple algorithm is used to extract the associated 25 rotation angle, which is the solid angle of the enclosed region. Section 3 presents a method to compute the solid 26 angle of a circular or elliptical disc shaped detector with respect to a point source having arbitrary location using an 27 n-sided polygon of equal area. Section 4 presents a method to compute the average solid angle subtended by a 28 disk detector with respect to an arbitrary positioned line source. This method includes line sources that are coaxial 29 to the detector’s axis and to line sources having axes parallel to but not coaxial to the detector axis. Section 5 30 presents a method to compute the average solid angle and geometrical efficiency of a disk source and coaxial disk 31 detector. A solid angle reciprocity relationship is also introduced that is based on heat transfer considerations. 32 Section 6 presents a method to compute the average solid angle associated with a disk source and detector system 33 having parallel but offset axes. The first case involves an offset distance greater than the radius of the source disk 34 and the second case involves an offset distance less than the source disk radius. Section 7 presents numerical 35 results with a comparison to exact solid angles based on an analytical equation for a disk detector and a point 36 source on the disk axis. Results from the proposed method are also compared to various published results 37 involving disk shaped detectors with point, line and disk sources. Section 8 presents the conclusions. 38 39 2. Computing solid angles using Pivot Parameters 40

A solid angle is the area on the surface of a unit sphere that is naturally parameterized using spherical coordinates 41 in eq. (1), where the integration is over the enclosed region. The ϴ integration is performed to reduce the 2-42 dimensional integral to a 1-dimensional path integral of φ about the perimeter of the enclosed area in the 43 counterclockwise direction, as shown in eq. (1). 44

Ω = ∬sin(𝛳)d𝛳d𝜙 = −∮cos(𝛳)d𝜙 (1) 45

According to Ishlinskii’s theorem, Ishlinskii (1952), the solid angle can also be obtained by slewing an axis around 46 the enclosed region on the unit sphere and computing the rotation angle associated with the slewing 47 transformation. In this application of Ishlinskii’s theorem, it should be noted that there is no component of 48 angular rate along the axis as it slews about the solid angle region. At the completion of the slewing motion, the 49 final orientation of the axis will be exactly the same as its initial orientation before the slewing motion. However, 50

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there will be a residual rotation about the axis that is given by an angle equal to the solid angle of the enclosed 1 area on the unit sphere. It should be noted that this angle, which is generated by attitude kinematics, is not 2 measurable by rate gyros. This new method of computing solid angle changes the problem from computing the 3 area on a unit sphere to that of computing the attitude transformation of an axis that slews about the solid angle 4 region on a unit sphere. The solid angle is easily obtained from its associated attitude transformation. The slewing 5 motion can be a finite number of great circle arcs, or and infinite number of infinitesimal great circle arc segments. 6 For solid angles associated with spherical polygons, a finite number of great circle arcs can be used to obtain the 7 exact solid angle value. For solid angle regions bounded by smooth arcs on the unit sphere, an infinite number of 8 infinitesimal great circle arc segments would be needed to obtain the exact solid angle value. However, the solid 9 angle can be approximated to any degree of accuracy using a large number of great circle arcs. 10

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Fig. 1. Spherical triangle defined by points a, b, c on the surface of the unit sphere is shown with the midpoints d, e, 21 f of the great circle are segments connecting the points. 22

Consider the solid angle of a spherical triangle defined by points a, b, c located on the surface of a unit sphere 23 shown in Fig. 1. An axis with one end located at the origin of the sphere moves its tip from a to b to c back to a, 24 thereby completing one circuit about the spherical triangle. The first great circle arc segment links point a to point 25 b, and has an associated attitude transformation given by U(a, b). The Euler rotation vector associated with this 26 transformation is directed normal to the great circle arc and has no component along the axis at any orientation of 27 the axis during its slewing motion. Slewing motion along a single great circle arc does not produce rotation about 28 the axis. The transformation produced by the axis after it has moved about the spherical triangle is given by the 29 product of the transformations associated with the three great circle arc segments in sequence, as shown in eq. 30 (2). 31

𝑼 = 𝑼(𝒂,𝒃)𝑼(𝒃, 𝒄)𝑼(𝒄, 𝒂) (2) 32

The attitude transformation associated with each great circle arc segment can be readily computed using the Pivot 33 Parameter formulation, Patera (2017). In the Pivot Parameter method, a rotation is specified by two Pivot 34 Vectors located in the rotation plane centered at the origin of the coordinate system. The angle between the Pivot 35 Vectors is one half of the desired rotation angle. A rotation of 180 degrees about the first Pivot Vector followed by 36 a rotation of 180 degrees about the second Pivot Vector achieves the desired rotation, which propagates each 37 Pivot Vector along a segment of a great circle arc that represents the desired rotation. Let d, e, and f be the 38 midpoints of segments a-b, b-c, and c-a, respectively. A 180 degree rotation about a followed by a 180 degree 39 rotation about d results in the desired great circle arc slewing from a to b, as shown in eq. (3). The same 40

A B

C

d

e

f

.

.

.

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transformation can be achieved using a 180 degree rotation about d followed by a 180 rotation about b, as shown 1 in eq. (3). 2

𝑼(𝒂, 𝒃) = 𝑹(𝒂, 𝟏𝟖𝟎)𝑹(𝒅, 𝟏𝟖𝟎) = 𝑹(𝒅, 𝟏𝟖𝟎)𝑹(𝒃, 𝟏𝟖𝟎) (3) 3

The rotation transformation defined by a rotation about r of angle α is given by the direction cosine matrix in eq. 4 (4), Patera (2010), where C=cos(α)-1 and S=sin(α). 5

R(𝐫, 𝛂) = (

1 + (rz2+ ry

2)C −(rxryC + rzS) ryS − rxrzC

rzS − rxryC 1 + (rx2 + rz

2)C −(rzryC + rxS)

−(rxrzC + ryS) rxS − ryrzC 1 + (rx2 + ry

2)C

) (4) 6

When α=180 degrees, eq. (4) reduces to eq. (5), which defines ψ(r) and R(r, 180). Note that the transformation is 7 orthonormal, since r is a unit vector. 8

𝐑(𝐫, 𝟏𝟖𝟎) = 𝛙(𝐫) = (

2rx2 − 1 2rxry 2rxrz

2rxry 2ry2 − 1 2𝑟𝑧𝑟𝑦

2rxrz 2𝑟𝑧𝑟𝑦 2rz2 − 1

) (5) 9

Using eq. (5) in eq. (3), one obtains eq. (6). 10

𝑼(𝒂, 𝒃) = 𝝍(𝒂)𝝍(𝒅) = 𝝍(𝒅)𝝍(𝒃) (6) 11

The same procedure can be used to obtain the other two great circle arc transformations given by eqs. (7) & (8). 12

𝑼(𝒃, 𝒄) = 𝝍(𝒃)𝝍(𝒆) = 𝝍(𝒆)𝝍(𝒄) (7) 13

𝑼(𝒄, 𝒂) = 𝝍(𝒄)𝝍(𝒇) = 𝝍(𝒇)𝝍(𝒂) (8) 14

Two rotations can be combined by joining their associated great circle arc segments at their intersection points. 15 Using eqs. (6) & (7), one finds the combined transformation from a to c, as shown in eq. (9). Notice that the two 16 180 rotations about the axis at point b results in a 360 degree rotation or no rotation at all. 17

𝑼(𝒂, 𝒃)𝑼(𝒃, 𝒄) = 𝝍(𝒅)𝝍(𝒃)𝝍(𝒃)𝝍(𝒆) = 𝝍(𝒅)𝝍(𝒆) (9) 18

The complete transformation around the spherical triangle, as given by eq. (2) can be obtained using eqs. (8) & (9), 19 as shown in eq. (10). 20

𝑼 = 𝑼(𝒂,𝒃)𝑼(𝒃, 𝒄)𝑼(𝒄, 𝒂) = 𝝍(𝒅)𝝍(𝒆)𝝍(𝒇)𝝍(𝒂) (10) 21

A similar equation can be created for a spherical quadrilateral with the midpoint of the fourth side given by g, as 22 shown in eq. (11). 23

𝑼 = 𝝍(𝒅)𝝍(𝒆)𝝍(𝒇)𝝍(𝒈) (11) 24

For any polygon having an even number of sides, the close loop transformation of the axis around the solid angle 25 region is given by the product of the transformations corresponding to 180 degree rotations about the midpoints 26 of the sides in the sequence corresponding to the axis trajectory. If the number of sides of the spherical polygon is 27 odd, an additional final rotation of 180 degrees about the initial point is required to complete the transformation, 28 as is shown in eq. (10). This method can be extended to any number of connected great circle arc segments that 29 form enclosed regions on the unit sphere. The solid angle can also be found for any region on the surface of a unit 30 sphere enclosed by a continuous curve, since the curve can be approximated to any degree of accuracy with a 31 large number of infinitesimal great circle arc segments. The attitude transformation associated with the great 32 circle arc segments can be computed using Pivot Parameters, as shown, and the associated rotation angle or solid 33 angle, can be extracted from the direction cosine matrix. Alternatively, the solid angle can be found as the 34

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magnitude of the Euler rotation vector corresponding to the transformation. Note that the Euler rotation vector 1 aligns with the initial orientation of the axis. 2

The rotation angle can be extracted from the direction cosine matrix by first computing the trace of the matrix, 3 which is the sum of the diagonal elements, as shown in eq. (12). 4

𝑇𝑟𝑎𝑐𝑒 = 𝑈(1,1) + 𝑈(2,2) + 𝑈(3,3) (12) 5

The solid angle, Ω, can be found using eq. (13). 6

𝛺 = 𝑐𝑜𝑠−1[𝑡𝑟𝑎𝑐𝑒−1

2] (13) 7

The components of the Euler rotation vector, V, can be found using eq. (14), once Ω is obtained from eq. (13). 8

𝑉(1) = [𝑈(3,2) − 𝑈(2,3)]𝛺

2sin(𝛺) 9

𝑉(2) = [𝑈(1,3) − 𝑈(3,1)]𝛺

2sin(𝛺) (14) 10

𝑉(3) = [𝑈(2,1) − 𝑈(1,2)]𝛺

2sin(𝛺) 11

Since the Ω can be as large as 2π but the largest magnitude of Ω that can be obtained by eq. (13) is only π, the sign 12 of V(3) can be used to determine the true value of Ω. If V(3) < 0, the solid angle is given by eq. (13). If V(3) > 0, the 13 solid angle is given by eq. (15). This logic depends on the details of the geometry. 14

𝛺 = 2𝜋 − 𝑐𝑜𝑠−1[𝑡𝑟𝑎𝑐𝑒−1

2] (15) 15

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3. The solid angle subtended by a circular or elliptical disk 18

The first application of the general method of computing solid angle is that of a point source and a circular disk 19 detector. It should be noted that Garner and Carnesale (1969), addressed this problem by replacing the circular 20 disk with a square of equal area. Their method worked well but required use of a correction factor table to be 21 accurate in cases of interest. The method was expanded by Garner and Verghese (1971), who replaced the circular 22 disk with an n-sided regular polygon of area equal to that of the disk. They found that a 36-sided regular polygon 23 was sufficient to obtain accurate results. The general method of computing solid angle, proposed in this work, also 24 replaces the circular disk with an n-sided regular polygon. The first step is to replace the radius of the disk with a 25 larger disk that encompasses the n-sided regular polygon of equal area. If R0 is the radius of the original disk, then 26 the radius of the enlarged disk is given by eq. (16), where n is the number of sides selected. 27

𝑅 = 𝑅0√[𝜋

𝑛𝑠𝑖𝑛(𝜋

𝑛)cos(

𝜋

𝑛)] (16) 28

The regular polygon is defined by vertices ordered in the counterclockwise direction and labeled by index k that 29 ranges from 1 to n+1. Let a(k) be the x component of the kth vertex and b(k) the y component of the kth vertex, as 30 shown in eq. (17). Notice that a(1) = a(n+1) and b(1) = b(n+1). 31

𝑎(𝑘) = 𝑅𝑐𝑜𝑠 [2𝜋(𝑘−1)

𝑛] , 𝑏(𝑘) = 𝑅𝑠𝑖𝑛 [

2𝜋(𝑘−1)

𝑛] (17) 32

Adjacent vertices are connected by straight line segments in the x-y plane. The midpoints of the line segments are 33 given by eq. (18), where m ranges from 1 to n. Note that z(m) = 0 for all m, since the polygon is in the x-y plane. 34

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𝑥(𝑚) = [𝑎(𝑚 + 1) + 𝑎(𝑚)]/2 , 𝑦(𝑚) = [𝑏(𝑚 + 1) + 𝑏(𝑚)]/2 (18) 1

The source, s, is placed anywhere in the upper hemisphere for convenience so that sequential axis locations are 2 defined by eq. (19), where r(m, i) defines the coordinates of the tip of the line segment with respect to the source. 3 Note that there is no restriction of the location of s, such as, being on the axis of the disk. 4

𝑟(𝑚, 1) = 𝑥(𝑚) − 𝑠(1) , 𝑟(𝑚, 2) = 𝑦(𝑚) − 𝑠(2) , 𝑟(𝑚, 3) = −𝑠(3) (19) 5

For simplicity, we choose n to be an even number, which results in an even number of axis orientations that define 6 the midpoint of the n-line segments. Each vector, r(m), defines the location of the mth midpoint with respect to 7 the source and must be unitized before it can be use in eq. (5). If n = 4, the attitude transformation associated 8 with the axis slewing about the 4-sided regular polygon is given by eq. (20), where r(m) are the unit vectors. 9

𝑼 = 𝝍[𝒓(𝟏)]𝝍[𝒓(𝟐)]𝝍[𝒓(𝟑)]𝝍[𝒓(𝟒)] (20) 10

Eq. (20) can be generalized to an n-sided regular polygon, as given by eq. (21). 11

𝑼 = ∏ 𝝍[𝒓(𝒎)]𝑛𝑚=1 (21) 12

Once the transformation is obtained for the closed path slewing motion of the axis about the solid angle region, 13 eqs. (13) and (14) are used to obtain the value of the solid angle. Note that although the polygon is located in the 14 x-y plane, the normalized vertices are located on the surface of a unit sphere centered at the source. The solid 15 angle region is the area on the unit sphere defined by the normalized vectors that define the vertices on the unit 16 sphere. Eq. (21) is the exact transformation associated with the slewing motion of the axis about the polygon 17 region. By choosing a large enough n, eq. (21) approximates the transformation of the axis about the original disk 18 in the x-y plane to any desired degree of accuracy within the limits of numerical processing errors. 19

The solid angle of an elliptical disk can be obtained by adjusting the x and y components in eq. (17). If the semi-20 major axis and semi-minor axis are given by A and B respectively, then eq. (17) becomes eq. (22). 21

𝑎(𝑘) = 𝐴𝑐𝑜𝑠 [2𝜋(𝑘−1)

𝑛] , 𝑏(𝑘) = 𝐵𝑠𝑖𝑛 [

2𝜋(𝑘−1)

𝑛] (22) 22

With the eq. (22) replacing eq. (17), the processing to compute the solid angle is exactly the same as for the 23 circular disk. This method does not require the source to be coaxial with the disk as does Thabet, et. al. (2020), 24 who computed the solid angle of a point source and a disk detector by transforming to axial geometry with an 25 equivalent elliptical detector cross section. The general method of computing solid angle applies to any location of 26 the source, as long as eq. (22) is used in place of eq. (17). 27

4. Average solid angle subtended by a disk detector with respect to an arbitrary linear source 28

The general method to compute solid angle developed in Section 3 can be used to find the average solid angle for 29 a disk detector with respect to a linear source. The linear source is represented by a discrete number of point 30 sources distributed uniformly along its length. Let the linear source extend from L1(i) to L2(i), where the index, i, 31 represents the x, y, and z components of the location. The linear source is represented by n point sources 32 separated by dL, as given in eq. (23). The location of each point source is given by eq. (24) to provide a uniform 33 distribution across the length of the source. 34

𝒅𝑳 = (𝑳𝟐 − 𝑳𝟏)/𝑛 (23) 35

𝑺(𝒌) = 𝑳𝟏 + 𝑘𝒅𝑳 − 𝒅𝑳/2 (24) 36

The kth point source location is used in the algorithm presented in Section 3 to generate the associated solid angle, 37 G(k), as given in eq. (25). 38

𝐺(𝑘) = 𝛺[𝑺((𝒌)] (25) 39

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The average solid angle for the linear source can be found by simply averaging the values of G(k), as shown in eq. 1 (26). 2

𝛺 =1

𝑛∑ 𝐺(𝑘)𝑛𝑘=1 (26) 3

Since, the orientation of the linear source does not have to be coaxial to the detector axis or even parallel to the 4 detector axis, eq. (26) is a general result. 5

The numerical integration method represented in eq. (26) can be enhanced to improve accuracy by employing a 6 two point Gaussian integration algorithm. In the Gaussian integration algorithm, eq. (24) is replaced with eq. (27), 7 where the single evaluation of each S(k) is replaced by two evaluations, S1(k) and S2(k). 8

𝑺𝟏(𝒌) = 𝑳𝟏 + 𝑘𝒅𝑳 −𝒅𝑳(1+1/√3)

2 , 𝑺𝟏(𝒌) = 𝑳𝟏 + 𝑘𝒅𝑳 −

𝒅𝑳(1−1/√3)

2 (27) 9

In a similar fashion, G in eq. (25) was replaced by G1 and G2 in eq. (28). 10

𝐺1(𝑘) = 𝛺[𝑺𝟏((𝒌)] , 𝐺2(𝑘) = 𝛺[𝑺𝟐((𝒌)] (28) 11

The average solid angle is obtained by replacing eq. (26) with eq. (29). 12

𝛺 =1

2𝑛∑ [𝐺1(𝑘) + 𝐺2(𝑘)]𝑛𝑘=1 (29) 13

The associated geometrical efficiency, GE, which is the solid angle divided by 4π, is given by eq. (30). 14

𝐺𝐸 = 𝛺/4𝜋 =1

8𝑛𝜋∑ [𝐺1(𝑘) + 𝐺2(𝑘)]𝑛𝑘=1 (30) 15

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5. Average solid angle of a disk detector with respect to a coaxial disk source 17

The average solid angle subtended by a coaxial disk detector with respect to a coaxial disk source can be found 18 using the results of Section 4, where the linear source is replaced with radial points that extend from the center of 19 the disk source to a point on the perimeter of the disk source. In this case, L1 is at the center of the disk source 20 and L2 is at any point on the perimeter of the disk source. The solid angle for the kth point between L1 and L2 is 21 given by eq. (25). The average solid angle for the disk source can be found by the weighted average of G(k), as 22 given in eq. (31), where dL is the magnitude of dL, as given by eq. (23). 23

𝛺 =2

(𝑳𝟐−𝑳𝟏)2∑ 𝐺(𝑘)𝑛𝑘=1 |𝑘𝒅𝑳 − 𝒅𝑳/2|𝑑𝐿 (31) 24

Note that eq. (31) can also be written, as shown in eq. (32). 25

𝛺 =2

(𝑳𝟐−𝑳𝟏)2∑ 𝐺(𝑘)𝑛𝑘=1 |𝑺(𝒌) − 𝑳𝟏|𝑑𝐿 (32) 26

Eq. (31) can be understood more easily by considering the average solid angle subtended by the disk detector over 27 the disk source using a continuous integral average, as shown in eq. (33), where G(r) is the associated continuous 28 function and ρ is the radius of the disk source. Notice that ρ is given by the absolute value of L2 - L1. 29

𝛺 = 2/𝜌2 ∫ 𝐺(𝑟)𝑟𝑑𝑟𝜌

0 (33) 30

The geometrical efficiency corresponding to eq. (33) is given by eq. (34). 31

𝐺𝐸 = 𝛺/4𝜋 =1

2𝜋𝜌2∫ 𝐺(𝑟)𝑟𝑑𝑟𝜌

0 (34) 32

9

The numerical integration method in eq. (31) was made more accurate by employing a two point Gaussian 1 integration algorithm, as presented in Section 4. As before, S1 and S2 are defined by eq. (27) and the respective 2 solid angles are defined by eq. (28). The average solid angle can be found by replacing eq. (32) with eq. (35). 3

𝛺 =1

(𝑳𝟐−𝑳𝟏)2∑ [𝐺1(𝑘)|𝑺𝟏(𝒌) − 𝑳𝟏| + 𝐺2(𝑘)|𝑺𝟐(𝒌) − 𝑳𝟏|]𝑑𝐿𝑛𝑘=1 (35) 4

Consider the case where the source and the disk are both black body radiators, which having emissivity and 5 absorptivity values of 1, and are in thermal equilibrium with each other. In this case, the infrared radiation 6 emitted by the source and absorbed by the detector is exactly equal to the infrared radiation emitted by the 7 detector and absorbed by the source. Since the total power radiated by source and detector is proportional the 8 respective surface areas, the heat balance equation between source and detector is given by eq. (36), where R is 9 the radius of the circular detector and ρ is the radius of the circular source. 10

𝜋𝜌2𝛺𝑠 = 𝜋𝑅2𝛺𝑅 (36) 11

The average solid angle of the detector as observed at the source is given by ΩS and the average solid angle of the 12 source as observed at the detector is given by ΩR. Therefore, the value of ΩR can be used to find the value of ΩS, as 13 shown in eq. (37). 14

𝛺𝑠 =𝑅2

𝜌2𝛺𝑅 (37) 15

This reciprocity relationship may be useful in cases where reversing the rolls of source and detector facilitate 16 computation or improve computation accuracy. 17

6. Average solid angle of a disk detector with respect to a parallel offset disk source 18

The average solid angle can be calculated for the case of a circular disk source and detector system having parallel 19 axes that are offset from coaxial alignment. The problem can be solved by considering a larger coaxial disk source 20 having a radius equal to the offset distance plus the radius of the original offset disk source, as shown in Fig. 2. 21

22

23

24

25

26

27

28

29

30

Fig. 2. Circular integration region when d is greater than ρ. 31

32

In this case, the integration region is not over the entire enlarged disk source, but just the original disk source. In 33 the situation shown in Fig. 2, the offset distance is larger than the radius of the source. The distance from the 34 detector axis in the x-y plane to the perimeter of the disk source is given by r, where r is related to d, ρ, and ϴ by 35 eq. (38.) 36

r

ρ

ϴ

d

10

𝑟2 = [𝑑 + 𝜌𝑐𝑜𝑠(𝛳)]2 + 𝜌2𝑠𝑖𝑛2(𝛳) (38) 1

In eq. (38), d is the shift distance, ρ is the radius of the source disk and ϴ is the angle defined in Fig. 2. By taking 2 the derivative of eq. (38), one finds the relationship between dr and dϴ, as shown in eq. (39). 3

𝑟𝑑𝑟 = −𝜌𝑑𝑠𝑖𝑛(𝛳)𝑑𝛳 (39) 4

The integral of the solid angle over the disk source region parameterized by r makes use of the coaxial symmetry, 5 as shown in eq. (40). The integration parameter is changed from r to a ϴ using eq. (38) and eq. (39), as shown in 6 eq. (40). G(r) is a function of ϴ through eq. (38), which relates r and ϴ. 7

𝐼 = ∫ 2𝐺(𝑟)𝜙𝑟𝑑𝑟 = −2𝜌𝑑 ∫ 𝐺(𝛳)𝜙𝑠𝑖𝑛(𝛳)𝑑𝛳0

𝜋= 2𝜌𝑑 ∫ 𝐺(𝛳)𝜙𝑠𝑖𝑛(𝛳)𝑑𝛳

𝜋

0

𝑑+𝜌

𝑑−𝜌 (40) 8

Note that φ is a function of ϴ, since, at each value of ϴ, φ is obtained as the inverse of tan(φ), as given by eq. (41). 9 Since d is greater than ρ, the value of φ is between 0 and 90 degrees. 10

𝜙 = 𝑡𝑎𝑛−1 [𝜌𝑠𝑖𝑛(𝛳)

𝑑+𝜌𝑐𝑜𝑠(𝛳)] (41) 11

Eq. (40) can be integrated numerically using the two-point gaussian algorithm with Δϴ=π/n and ϴ1 and ϴ2 defined 12 in eq. (42). 13

𝛳1(𝑘) = 𝑘𝛥𝛳 −𝜟𝛳(1+1/√3)

2 , 𝛳2(𝑘) = 𝑘𝛥𝛳 −

𝜟𝛳(1−1/√3)

2 (42) 14

Using the definitions in eqs. (43) - (45), one finds the numerical integral of eq. (40), as shown in eq. (46). 15

𝐺1(𝑘) = 𝛺[𝛳1(𝑘)] , 𝐺2(𝑘) = 𝛺[𝛳2(𝑘)] (43) 16

𝜙1(𝑘) = 𝜙[𝛳1(𝑘)] , 𝜙2(𝑘) = 𝜙[𝛳2(𝑘)] (44) 17

𝑠𝑖𝑛1(𝑘) = 𝑠𝑖𝑛[𝛳1(𝑘)] , 𝑠𝑖𝑛2(𝑘) = 𝑠𝑖𝑛[𝛳2(𝑘)] (45) 18

𝐼 =𝜌𝑑

𝑛∑ [𝐺1(𝑘)𝜙1(𝑘)𝑠𝑖𝑛1(𝑘) + 𝐺2(𝑘)𝜙2(𝑘)𝑠𝑖𝑛2(𝑘)]𝛥𝛳𝑛𝑘=1 (46) 19

The average solid angle over the disk source for the continuous integral is found by dividing eq. (40) by the area of 20 the disk source as given by eq. (47). Therefore, the numerical summation in eq. (46) must be divided by the area 21 of the disk source to obtain the average solid angle, as given by eq. (48). 22

< 𝛺 >=𝐼

𝜋𝜌2=

2𝜌𝑑 ∫ 𝐺𝜙𝑠𝑖𝑛(𝛳)𝑑𝛳𝜋0

𝜋𝜌2 (47) 23

< 𝛺 >=𝑑

𝜋𝑛𝜌∑ [𝐺1(𝑘)𝜙1(𝑘)𝑠𝑖𝑛1(𝑘) + 𝐺2(𝑘)𝜙2(𝑘)𝑠𝑖𝑛2(𝑘)]𝛥𝛳𝑛𝑘=1 (48) 24

The second situation shown in Fig. 3, has the offset distance smaller than the radius of the disk source. Once 25 again, we consider a larger coaxial disk source having radius equal to d plus ρ. In this case, eq. (38) and eq. (39) are 26 replaced with eq. (49) and eq. (50), respectively. 27

𝑟2 = [𝜌𝑐𝑜𝑠(𝛳) − 𝑑]2 + 𝜌2𝑠𝑖𝑛2(𝛳) (49) 28

𝑟𝑑𝑟 = 𝜌𝑑𝑠𝑖𝑛(𝛳)𝑑𝛳 (50) 29

The integral of the enlarged coaxial disk source, IE, minus the original shifted disk source, IS, is given by I, as shown 30 in eq. (51), where eq. (50) has been used to change the integration parameter to ϴ. 31

𝐼 = 𝐼𝐸 − 𝐼𝑆 = ∫ 2𝐺𝜙𝑟𝑑𝑟 = 2𝜌𝑑 ∫ 𝐺𝜙𝑠𝑖𝑛(𝛳)𝑑𝛳𝜋

0

𝑑+𝜌

𝜌−𝑑 (51) 32

11

Although eq. (51) has the same form as eq. (40), the integrand is different, since eq. (41) is replaced by eq. (52), 1 when determining φ from ϴ. Since ρ is greater than d in eq. (52), tan(φ) can be negative and the value of φ is 2 between 0 and 180 degrees. In addition, the replacement of eq. (38) with eq. (49) changes the value of G in the 3 integrand. 4

𝜙 = 𝑡𝑎𝑛−1 [𝜌𝑠𝑖𝑛(𝛳)

𝜌𝑐𝑜𝑠(𝛳)−𝑑] (52) 5

Eq. (51) is evaluated numerically using the two point gaussian integration algorithm, as given by eq. (46) but with 6 the appropriate changes to the integrand. IE is evaluated using the method of Section 5 and is given by eq. (53). 7

𝐼𝐸 = 2𝜋 ∫ 𝐺𝑟𝑑𝑟𝜌+𝑑

0 (53) 8

Once values for I and IE are obtained from eq. (51) and eq. (53), the average solid angle over the shifted disk source 9 is obtained by eq. (54). Therefore, one doesn’t evaluate the average solid angle over the shifted source directly, 10 since it is easily obtained indirectly using I and IE. The related geometrical efficiency is obtained by dividing eq. (54) 11 by 4π, as given by eq. (55). 12

< 𝛺 >=𝐼𝑆

𝜋𝜌2=

𝐼𝐸−𝐼

𝜋𝜌2 (54) 13

𝐺𝐸 =𝐼𝑆

4𝜋2𝜌2=

𝐼𝐸−𝐼

4𝜋2𝜌2 (55) 14

15

16

17

18

19

20

21

22

23

24

25

26

27

Fig. 3. Integration region is between the circles when d is less than ρ. 28

7. Numerical results 29

The new method to compute solid angle was implemented in several computer simulations to demonstrate its 30 performance. Results of the new method were first compared to the exact solution of a simple known case 31 involving a point source located on the axis of a disk detector. For a point source located at a distance h along the 32 axis of a disk having radius R, the solid angle is given by eq. (56). 33

d

r φ

ϴ

ρ

ρ+d

12

𝛺 = 2𝜋 [1 −ℎ

√𝑅2+ℎ2] (56) 1

Figure 4 shows the solid angle obtained using eq. (56), when R = 10 and h ranges from 0 to 200. The solid angle 2 was also computed using the general method with a 360-sided regular polygon in place of the disk. The polygon 3 was expanded to have an area equal to that of the disk. Figure 5 shows the percent difference between the solid 4 angle of the polygon, which was found using the general method, and the solid angle of the disk using eq. (56). 5 Note that the solid angle of the polygon is exact except for numerical processing errors. Therefore, the error in Fig. 6 5 shows how well the polygon’s solid angle represents that of the disk. When the number of sides of the polygon 7 was increased from 360 to 800, it more accurately represented the circular disk and the percent error decreased, 8 as shown in Fig. 5. The noisy values in Fig. 5 are caused by numerical processing error at larger values of h. The 9 extremely small error displayed in Fig. 5 indicates that the general method using the polygon is highly accurate. 10 The advantage of the general method is that the point source can be located at any point and does not have to be 11 on the axis of the disk, as required by eq. (56). 12

13

14

15

Fig. 4. Exact solid angle of a disk having radius or 10 meters with respect to a point above a disk along its axis. 16

17

18

19

20

21

22

0.1

1

10

100

1000

0 50 100 150 200 250

Solid

An

gle,

deg

s.

Height Above Disk, m

13

1

Fig. 5. Percent error of the general method when using 360-sided and 800-sided polygons to represent the 2 circular disk. 3

4

7.1 Point source on the axis of an elliptical disk detector 5

The general method can be modified easily to compute the solid angle of an elliptical disk with respect to a point 6 source along the axis normal to the plane of the disk, as presented in Section 3. Let A and B be the semi-major axis 7 and semi-minor axes of the ellipse, respectively. In this case, eq. (17) is replaced with eq. (22) and the remainder 8 of the processing is the same. 9

Table 1. Solid angle comparison for an elliptical detector with point source on the detector axis 10

Case #

A B h Conway (2010a)

Thabet, et. al. (2020)

General Method (n=360)

1 2 1 0.01 6.2347447 6.2346390 6.2347447

2 2 1 0.5 4.0425592 4.0423850 4.0425592

3 2 1 10 0.0616819 0.0616804 0.0616819

4 1000 1 1 3.1415778 3.1414940 3.1416101 [3.1415778, (n = 2200)]

5 100 1 1 3.1405637 3.1404830 3.1405643 [3.1405637, (n = 500)]

6 1 0.99 1 1.8291325 1.8290270 1.8291325

7 1 0.2 1 0.4724316 0.4724257 0.4724316

8 1 0.01 1 0.0239619 0.0239632 0.0239619

9 1 0.001 1 0.0023963 0.0023972 0.0023963

11

The results using the general method was compared to published results, Thabet, et. al. (2020) in Table 1, which 12 includes results from Conway (2010a). The results of the proposed general method, when rounded down to seven 13 digits to the right of the decimal point, agreed with the results of Conway (2010a). All cases used a 360-sided 14 polygon. The number of sides for case #4 was increased to 2200, as shown in parentheses, to match Conway’s 15 result exactly. The number of sides for case #5 was increased to 500 to match Conway’s result exactly. Although 16 cases #4 and #5 are numerically stressful cases, the general method proposed in this work can achieve any desired 17 degree of accuracy by increasing the number of sides of the polygon. The only limitation is numerical processing 18 error. 19

The general method can also handle cases where the point source is not located on the axis of the elliptical 20 detector. Table 2 shows that the general method, while using a 360-sided polygon, produces results in exact 21

1.0E-11

1.0E-10

1.0E-09

1.0E-08

1.0E-07

1.0E-06

1.0E-05

0 50 100 150 200 250

Pe

rce

nt

Err

or

Height Above Disk, m

n=800 n=360

14

agreement to those of Conway (2010a). The results in Table 2 show that the general method is valid for sources 1 located anywhere with respect to the elliptical detector. 2

Table 2. Solid angle for an elliptical detector with respect to a point source located at (p, q, h) 3

Case # A B p q h Conway (2010a) General Method (n=360)

1 1.5 1.0 0.6 0.5 0.1 5.4932906 5.4932906

2 1.5 1.0 0.6 0.5 2.52 0.5690505 0.5690505

3 1.5 1.0 0.6 0.5 10.0 0.0461472 0.0461472

4 1.0 4.0 2.0 0.5 0.1 0.1104206 0.1104206

5 1.0 4.0 0.1 4.0 5.0 0.2589691 0.2589691

6 2.0 1.0 1.0 0.5 1.0 1.9450789 1.9450789

7 2.0 1.0 5.0 1.0 1.0 0.0547517 0.0547517

8 3.0 1.0 1.0 0.5 0.5 3.6792113 3.6792113

9 5.0 1.0 1.0 0.5 0.5 3.9166119 3.9166119

10 1.1 1 1 0.9 2.0 0.4744215 0.4744215

4

7.2 Line source parallel to the axis of a circular disk detector 5

The average solid angle subtended by a disk detector with respect to a line source parallel to the disk axis was 6 computed using the method of Section 4. In Section 4 the points were distributed uniformly over the length of the 7 line source and the associated solid angles were averaged to obtain the average solid angle subtended by the disk 8 detector. The method was then enhanced to improve accuracy by employing a two point Gaussian integration 9 algorithm. The algorithm was applied to computing the geometrical efficiency of a line source along the axis of a 10 disk detector and the results were compared to similar results published by Abbas (2018), in Table 3, which also 11 contains results from Conway (2010b). 12

Table 3. Geometrical efficiency for a coaxial line source of length 2a 13

Case #

R h a Conway (2010b) Abbas (2018) General Method (n=360)

1 10 5 3 0.2826492 0.2826492 0.2826492

2 10 5 5 0.2928932 0.2928932 0.2928932

3 105 5 5 0.4999750 0.4999750 0.4999750

4 10 1 1 0.4504902 0.4504902 0.4504902

5 10 0.1 0.1 0.4950005 0.4950005 0.4950005

6 10 100 2 0.0024824 0.0024824 0.0024824

7 10 100 100 0.0243754 0.0243754 0.0243754

8 5 5 5 0.1909830 0.1909830 0.1909830

9 10 10 2 0.1482186 0.1482186 0.1482186

14

In Table 3, R is the radius of the circular disk detector, h is the distance along the detector axis corresponding to 15 the midpoint of the line source and a is one half of the length of the line source. All the results were obtained 16 using a 360-sided polygon and were rounded down to seven digits to the right of the decimal point to provide 17 uniformity, although the general method agreed with the published results to many more digits. 18

The general method was also applied to computing the geometrical efficiency of a line source parallel to the axis of 19 the circular disk detector, but not coaxial to the disk axis. The results were in perfect agreement with similar 20 published results of Abbas (2018), and Conway (2010b), as shown in Table 4. The notation in Table 4 is the same 21

15

as that of Table 3 but Table 4 has the additional parameter, d, which is the distance that the parallel line source 1 was shifted from the axis of the disk detector. All results were rounded down to seven digits to the right of the 2 decimal to maintain uniformity. 3

Table 4. Geometrical efficiency for a parallel line source shifted by d and of length 2a 4

Case # R d h a Conway (2010b) Abbas (2018) General Method (n=360)

1 10 1 6 1 0.2424075 0.2424075 0.2424075

2 10 1 6 6 0.2650418 0.2650418 0.2650418

3 50 1 6 6 0.4408232 0.4408232 0.4408232

4 1000 1 6 6 0.4970001 0.4970001 0.4970001

5 5 1 5 4 0.1726042 0.1726042 0.1726042

6 5 10 5 4 0.0226308 0.0226308 0.0226308

7 20 5 1 1 0.4738339 0.4738339 0.4738339

8 20 5 10 1 0.2697133 0.2697133 0.2697133

9 20 5 40 1 0.0519778 0.0519778 0.0519778

5

7.3 Circular disk source coaxial to a disk detector 6

The geometrical efficiency associated with a coaxial disk source with respect to a circular disk detector was 7 computed according to Section 5 and compared to published results Selim and Abbas (1994) in Table 5. 8

Table 5. Solid angle for coaxial circular detector of radius R separated by h from a circular source of radius ρ 9

Case # R h ρ Selim and Abbas (1994)

General Method (n=360)

1 1 10 1 0.03097 0.0309543

2 1 10 4 0.02792 0.0279254

3 1 10 7 0.02309 0.0230928

4 1 10 10 0.01836 0.0183614

5 12 10 12 1.75641 1.7564276

6 12 10 15 1.52114 1.5211577

7 12 10 17 1.36655 1.3665614 8 12 10 19 1.22193 1.2219275

9 28 10 23 3.60777 3.6078341

10 28 10 25 3.47289 3.4729432

11 28 10 28 3.23495 3.2349867

12 28 10 30 3.05770 3.0577027

10

Although the published results in Table 5 include only five digits to the right of the decimal point, the general 11 method included seven digits. It is evident in Table 5, that some of the published results are not as accurate as 12 that of the general method, which used a 360-sided polygon. If such accurate results are not essential, the number 13 of sides of the representative polygon can be reduced. 14

The general method was tested against results of a more recent publications, Vega-Carrillo (1996), Conway (2006), 15 and Thabet, et. al. (2020), as shown in Table 6. In this case, only four digits to the right of the decimal point were 16 included in the published results, except for Conway’s results, which included seven digits. Results of the general 17 method agree exactly with Conway’s results in all cases. Cases #6, #8, and #9 have values from the general 18 method that when rounded to only four digits agree with results of either Vega-Carrillo (1996), or Thabet, et. al. 19

16

(2020) or both. These results indicate that round-off differences may have contributed to the discrepancy in some 1 of the published results. 2

Table 6. Geometrical efficiency for a coaxial circular detector of radius R separated by h from a circular source of 3 radius ρ 4

Case # R h ρ Vega-Carrillo (1996) Thabet, et. al. (2020)

Conway (2006)

General Method (n=360)

1 1 1 1 0.1161 0.1165 0.1161054 0.1161054

2 4 1 1 0.3761 0.3774 0.3761319 0.3761319

3 1 5 1 0.0094 0.0095 0.0094456 0.0094456

4 8 5 1 0.2342 0.2350 0.2341957 0.2341957

5 2.54 5 0.3 0.0541 0.0543 0.0541240 0.0541240

6 2.54 20 0.3 0.0040 0.0040 0.0039835 0.0039835

7 2.54 5 2 0.0501 0.0503 0.0501442 0.0501442

8 2.54 10 2 0.0150 0.0150 0.0149857 0.0149857

9 2.54 20 2 0.0039 0.0040 0.0039553 0.0039553

5

7.4. Disk detector with respect to a parallel offset disk source 6

The geometrical efficiency associated with a disk source whose axis is parallel to the axis of a disk detector was 7 computed according to Section 6 and compared to published results, Conway (2006) in Table 7. The circular disk 8 detector was modeled as a 360-sided polygon and the angular integration in eq. (40) and eq. (45) used 800 steps 9 with a two point gaussian integration scheme. Only ten digits to the right of the decimal point were retained in 10 Table 7, although in most cases agreement between the results extended to several more digits. Results from the 11 general method matched Conway’s results, in all cases except case #8. When the number of sides of the polygon 12 was increased from 360 to 800 for case #8, results from the general method matched Conway’s results to 11 digits 13 to the right of the decimal point, as shown in Case #8a. The agreement of the results in Table 7 shows that the 14 accuracy of the general method is only limited by the number of sides of the representative polygon and numerical 15 processing error. 16

17

18

19

20

21

22

23

24

25

26

27

17

Table 7. Geometrical efficiency of a disk detector of radius R separated by h from a parallel disk source of radius ρ 1 offset by distance d 2

Case # ρ h R d Conway (2006) General Method

(360-sided polygon)

1 1.0 1.0 0.5 0 0.0345720168 0.0345720168

2 1.0 1.0 0.5 0.2 0.0339484847 0.0339484847

3 1.0 1.0 0.5 0.4 0.0321294982 0.0321294982

4 1.0 1.0 0.5 0.6 0.0292784973 0.0292784973

5 1.0 1.0 0.5 0.8 0.0256782337 0.0256782337

6 1.0 1.0 0.5 1.0 0.0217032254 0.0217032254

7 1.0 1.0 2.0 0 0.2620768484 0.2620768484

8 1.0 1.0 2.0 0.5 0.2539076886 0.2539076885

8a 1.0 1.0 2.0 0.5 0.25390768855 0.25390768855 (800-sided polygon)

9 1.0 1.0 2.0 1.0 0.2284050042 0.2284050042

10 1.0 1.0 2.0 2.0 0.1357136065 0.1357136065

11 1.0 1.0 2.0 5.0 0.0094641787 0.0094641787 12 1.0 1.0 2.0 10.0 0.0010424862 0.0010424862

13 1.0 5.0 2.0 1.0 0.0334471117 0.0334471117

14 1.0 5.0 2.0 5.0 0.0143513580 0.0143513580

15 1.0 5.0 2.0 10.0 0.0036868176 0.0036868176

16 1.0 5.0 2.0 20.0 0.0005775649 0.0005775649

3

Conclusion 4

A general method to compute solid angle was developed that uses the attitude transformation of an axis that 5 slews about a solid angle region. This method uses the results of Ishlinskii’s theorem to establish the relationship 6 between the solid angle and the region swept out by a slewing axis. Once the solid angle region is defined by a set 7 of points on its perimeter, the Pivot Parameter method is used to compute the transformation of the slewing 8 motion of the axis about the closed path. After the axis completes its closed path motion and returns to its initial 9 orientation, the resulting rotation about the axis is the solid angle of the enclosed region. The method was applied 10 to the solid angle of a disk detector with respect to a point source at an arbitrary location. Since the Pivot 11 Parameter method involves great circle arc slews that connect the points that define the solid angle region, the 12 circular disk was modeled as an n-sided polygon of area equal to that of the disk. The Pivot Parameter method 13 yields the exact solid angle of the polygon, which is a very good approximation to the solid angle of the disk. A 14 scale change applied to the polygon enables it to represent an elliptical shaped disk and permits the solid angle of 15 the elliptical region to be evaluated exactly like the circular disk. A line source was modeled by a number of point 16 sources distributed uniformly along the line. The average of the solid angles associated with the point sources 17 provided the average solid angle of the line source. The average solid angle of a disk detector-source system 18 having parallel axes was also obtained. Published test cases involving point, line and disk sources, were used to 19 help establish the accuracy of this new method. If higher accuracy is required, the number of sides of the 20 representative polygon can be increased such that the polygon represents the disk to a higher fidelity. 21 Computation effort is not great, since it requires only a single 180-degree rotation per side of the polygon to 22 obtain the associated attitude transformation of the axis. This new method is not limited to disk shaped sources, 23 since any source geometry can be represented by a distribution of source points and any detector shape can be 24 modeled as a set of perimeter points. Therefore, this method can treat any source-detector geometry desired. 25 The method can be implemented using the two point gaussian integration algorithm or any suitable software 26 integration package. A significant advantage of this new method is that it does not involve special mathematical 27

18

functions, integrals, or power series. This general solid angle computation method can be implemented easily in 1 computer simulations to compute solid angles that are needed in diverse disciplines, such as, radiant heat transfer, 2 particle scattering cross sections, electric and magnetic field strength, ligand size evaluation, rendering of 3-3 dimensional objects in computer graphics, as well as calibrating the strength of radioisotopes. Since the proposed 4 general method can treat any geometric configuration, it can be applied to future solid angle calculation problems 5 that are not yet defined. 6

7

References 8

Abbas, M. I., “Analytical Calculations of the Solid Angle Subtended by a Circular Detector at Linear Sources,” 9 Scientific Publication of the State University of Novi Pazar Ser. A: Apl. Math Inform. And Mech. 10, 1, 17-23, (2018). 10

Conway, J. T., “Generalizations of Ruby's formula for the geometric efficiency of a parallel-disk source and detector 11 system,” Nuclear Instruments and Methods Physics Research Section A: Accelerators, Spectrometers, Detectors, 12 and Associated Equipment 562, 146-153, (2006). 13

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